Let $G=\langle X;R\rangle$ be a finitely presented group. The rank of $G$ is defined to be the size of smallest generating set of $G$. The deficiency ${\rm def}(G)$ of $G$ is defined to be the maximum of $|X| - |R|$ over all finite presentations $G = \langle X;R \rangle$.

The deficiency of an amenable group can be at most $1$. One (maybe the only?) way to see this is to note that there is a Morse inequality for $\ell^2$-homology
$$1-{\rm def}(G) \geq b_0^{(2)}(G) - b_1^{(2)}(G) + b_2^{(2)}(G),$$
where $b_i^{(2)}(G)$ denotes the $i$-th $\ell^2$-Betti number of $G$. Cheeger and Gromov showed that an amenable group satisfies $b_i^{(2)}(G)=0$ for $i \geq 1$. This implies in particular that ${\rm def}(G) \leq 1$ for $G$ amenable.

Now, apart from $\mathbb Z$ and Baumslag-Solitar groups $BS(1,n) = \langle a,b; a^nba^{-1}b^{-1} \rangle$, I do not know of any amenable groups which realize ${\rm def}(G) =1$. In particular, I do not know any examples of rank $\geq 3$.

Question: Does every amenable group of deficiency $1$ have rank $\leq 2$.

It can be shown that any amenable group with ${\rm def}(G)=1$ must have cohomological dimension $\leq 2$, which puts severe restrictions on $G$. What else is known about amenable groups with deficiency $1$?

To see an amenable group can have deficiency at most 1, you can use the Baumslag-Pride theorem, which among other things implies a group with deficiency at least 2 has a free subgroup.
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Steve DApr 10 '12 at 17:47

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@Andreas: You can try to use Moldavansky's decomposition of 1-relator groups to show that 2-generator, 1-relator amenable groups are infinite cyclic or Baumslag-Solitar (Moldavansky gives you actions on simplicial trees, while amenable groups would have to fix points in the tree or at infinity of the tree). In general, I would try the technique of Baumslag-Shalen, using character varieties: The technique works for deficiency $\ge 2$, but might yield something interesting in the deficiency 1 case as well since amenability among linear groups is very rare.
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MishaApr 11 '12 at 14:26

@Misha: a 1-relator group having deficiency 1 has rank 2 (=1+1). Amenable 1-related groups are Baumslag-Solitar groups only (that includes $\mathbb{Z}^k, k=0,1,2$). It is well known.
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Mark SapirApr 11 '12 at 18:17

Mark, I was not sure about this. Do you have a reference?
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Andreas ThomApr 11 '12 at 19:57

@Andreas: This is stated in the paper by Grigorchuk and Ceccherini-Silberstein "Amenability and growth of one-relator groups". They do not prove the result but state that it follows from a paper by Karrass and Solitar. I assume that it does and the proof is probably by looking at the Moldavansky hierarchy or a similar consideration.
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MishaApr 11 '12 at 22:58

1 Answer
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Inspired by John Wilson's result which was mentioned by Mark in his answer (which he has deleted by now), I can answer my own question now for elementary amenable groups.

Let us first argue that any amenable $G$ with ${\rm def}(G)=1$ has cohomological dimension $\leq 2$. This follows from an argument that I learned from [1]. Consider the presentation $2$-complex $X$ (with one 0-cell) of a presentation which realizes the deficiency. The cellular chain complex of the universal covering looks like
$$0 \to {\mathbb Z}G^n \stackrel{d}\to {\mathbb Z}G^{n+1} \to {\mathbb Z}G \to 0.$$

The kernel of $d$ is $\pi_2(X)$ and the only thing to show is that $\pi_2(X)=0$. Then $X$ is aspherical and ${\rm cd}(G)\leq 2$. Now, the $\ell^2$-homology of $X$ is computed by
$$0 \to ({\ell^2}G)^n \stackrel{d}\to ({\ell^2}G)^{n+1} \to {\ell^2}G \to 0.$$
Now, the zeroeth and first $\ell^2$-Betti number of $X$ agrees with the $\ell^2$-Betti number of $G$ and has to vanish by Cheeger-Gromov since $G$ is amenable. This implies that $d$ must be injective on $(\ell^2 G)^n$. Hence, it is injective on $(\mathbb Z G)^n$. We conclude that $\pi_2(X)=0$.

Let us come to the main part of the argument. Jonathan Hillman has defined the Hirsch length $h(G)$ for any elementary amenable group $G$ (Theorem 1 in [3]) and shown that it is bounded above by the cohomological dimension of $G$, see Lemma 2 in [3]. Now, Theorem 2 of [3] implies $G/T$ is solvable where $T$ is the maximal locally finite normal subgroup of $G$. However, since the cohomological dimension is finite, $T$ is trivial and we conclude that $G$ itself is solvable. Now, Theorem 5 from [2] says that every solvable group of cohomological dimension $\leq 2$ must be solvable Baumslag-Solitar, $\mathbb Z^2$ or $\mathbb Z$.